Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $q \neq 0$. $r = \dfrac{-6}{8q + 40} \times \dfrac{7(q + 5)}{8q} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ -6 \times 7(q + 5) } { (8q + 40) \times 8q } $ $ r = \dfrac {-6 \times 7(q + 5)} {8q \times 8(q + 5)} $ $ r = \dfrac{-42(q + 5)}{64q(q + 5)} $ We can cancel the $q + 5$ so long as $q + 5 \neq 0$ Therefore $q \neq -5$ $r = \dfrac{-42 \cancel{(q + 5})}{64q \cancel{(q + 5)}} = -\dfrac{42}{64q} = -\dfrac{21}{32q} $